\(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x) \, dx\) [1369]

Optimal result

Integrand size = 35, antiderivative size = 209 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a b (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 \left (4 A b^2+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {8 a A b \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 d}+\frac {2 A (a+b \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:

-2/5*(5*b^2*(A-C)+a^2*(3*A+5*C))*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/ 
2*c),2^(1/2))*sec(d*x+c)^(1/2)/d+4/3*a*b*(A+3*C)*cos(d*x+c)^(1/2)*InverseJ 
acobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/d+2/5*(4*A*b^2+a^2*(3*A+5* 
C))*sec(d*x+c)^(1/2)*sin(d*x+c)/d+8/15*a*A*b*sec(d*x+c)^(3/2)*sin(d*x+c)/d 
+2/5*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 5.13 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.70 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sec ^{\frac {5}{2}}(c+d x) \left (-3 \left (5 b^2 (A-C)+a^2 (3 A+5 C)\right ) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 a b (A+3 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+3 \left (a^2 A+\left (5 A b^2+a^2 (3 A+5 C)\right ) \cos ^2(c+d x)\right ) \sin (c+d x)+5 a A b \sin (2 (c+d x))\right )}{15 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2) 
,x]
 

Output:

(2*Sec[c + d*x]^(5/2)*(-3*(5*b^2*(A - C) + a^2*(3*A + 5*C))*Cos[c + d*x]^( 
5/2)*EllipticE[(c + d*x)/2, 2] + 10*a*b*(A + 3*C)*Cos[c + d*x]^(5/2)*Ellip 
ticF[(c + d*x)/2, 2] + 3*(a^2*A + (5*A*b^2 + a^2*(3*A + 5*C))*Cos[c + d*x] 
^2)*Sin[c + d*x] + 5*a*A*b*Sin[2*(c + d*x)]))/(15*d)
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.93, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4709, 3042, 3527, 27, 3042, 3510, 27, 3042, 3500, 27, 3042, 3227, 3042, 3119, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x)^{7/2} (a+b \cos (c+d x))^2 \left (A+C \cos (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x))^2 \left (C \cos ^2(c+d x)+A\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\)

\(\Big \downarrow \) 3527

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{5} \int \frac {(a+b \cos (c+d x)) \left (-b (A-5 C) \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+4 A b\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {(a+b \cos (c+d x)) \left (-b (A-5 C) \cos ^2(c+d x)+a (3 A+5 C) \cos (c+d x)+4 A b\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3510

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int -\frac {-3 b^2 (A-5 C) \cos ^2(c+d x)+10 a b (A+3 C) \cos (c+d x)+3 \left ((3 A+5 C) a^2+4 A b^2\right )}{2 \cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (A-5 C) \cos ^2(c+d x)+10 a b (A+3 C) \cos (c+d x)+3 \left ((3 A+5 C) a^2+4 A b^2\right )}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {-3 b^2 (A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+10 a b (A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+5 C) a^2+4 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3500

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (2 \int \frac {10 a b (A+3 C)-3 \left ((3 A+5 C) a^2+5 b^2 (A-C)\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (\int \frac {10 a b (A+3 C)-3 \left ((3 A+5 C) a^2+5 b^2 (A-C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (\int \frac {10 a b (A+3 C)-3 \left ((3 A+5 C) a^2+5 b^2 (A-C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (-3 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) \int \sqrt {\cos (c+d x)}dx+10 a b (A+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (-3 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+10 a b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (10 a b (A+3 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (-\frac {6 \left (a^2 (3 A+5 C)+5 b^2 (A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {6 \left (a^2 (3 A+5 C)+4 A b^2\right ) \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {20 a b (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {8 a A b \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

Input:

Int[(a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + b*Cos[c + d*x])^2*Sin[c + 
 d*x])/(5*d*Cos[c + d*x]^(5/2)) + ((8*a*A*b*Sin[c + d*x])/(3*d*Cos[c + d*x 
]^(3/2)) + ((-6*(5*b^2*(A - C) + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2 
])/d + (20*a*b*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/d + (6*(4*A*b^2 + a^2* 
(3*A + 5*C))*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]))/3)/5)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

rule 3510

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ 
e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S 
imp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( 
m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m 
 + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) 
))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F 
reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 
 0] && LtQ[m, -1]
 

rule 3527

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + 
 f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 
2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* 
d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b 
*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( 
A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - 
b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [F(-1)]

Timed out.

hanged

Input:

int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)
 

Output:

int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.27 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {-10 i \, \sqrt {2} {\left (A + 3 \, C\right )} a b \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 10 i \, \sqrt {2} {\left (A + 3 \, C\right )} a b \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a^{2} + 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a^{2} - 5 i \, {\left (A - C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (10 \, A a b \cos \left (d x + c\right ) + 3 \, A a^{2} + 3 \, {\left ({\left (3 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algori 
thm="fricas")
 

Output:

1/15*(-10*I*sqrt(2)*(A + 3*C)*a*b*cos(d*x + c)^2*weierstrassPInverse(-4, 0 
, cos(d*x + c) + I*sin(d*x + c)) + 10*I*sqrt(2)*(A + 3*C)*a*b*cos(d*x + c) 
^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*( 
I*(3*A + 5*C)*a^2 + 5*I*(A - C)*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, 
 weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(- 
I*(3*A + 5*C)*a^2 - 5*I*(A - C)*b^2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, 
 weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(10*A*a*b* 
cos(d*x + c) + 3*A*a^2 + 3*((3*A + 5*C)*a^2 + 5*A*b^2)*cos(d*x + c)^2)*sin 
(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algori 
thm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(7/2) 
, x)
 

Giac [F]

\[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^2*sec(d*x + c)^(7/2) 
, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^2,x 
)
 

Output:

int((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(7/2)*(a + b*cos(c + d*x))^2, 
x)
 

Reduce [F]

\[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}d x \right ) a^{2} b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )^{3}d x \right ) b^{2} c +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{3}d x \right ) a b c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a^{2} c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3}d x \right ) a \,b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{3}d x \right ) a^{3} \] Input:

int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x)
 

Output:

2*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x)**3,x)*a**2*b + int(sqrt 
(sec(c + d*x))*cos(c + d*x)**4*sec(c + d*x)**3,x)*b**2*c + 2*int(sqrt(sec( 
c + d*x))*cos(c + d*x)**3*sec(c + d*x)**3,x)*a*b*c + int(sqrt(sec(c + d*x) 
)*cos(c + d*x)**2*sec(c + d*x)**3,x)*a**2*c + int(sqrt(sec(c + d*x))*cos(c 
 + d*x)**2*sec(c + d*x)**3,x)*a*b**2 + int(sqrt(sec(c + d*x))*sec(c + d*x) 
**3,x)*a**3